Let f = ln(x + √(x ² + y ²)) ln(sin(y/x)).
Then the total differential is
[tex]\mathrm df = \dfrac{\mathrm d\left(x+\sqrt{x^2+y^2}\right)}{x+\sqrt{x^2+y^2}}\ln\left(\sin\left(\dfrac yx\right)\right) + \ln\left(x+\sqrt{x^2+y^2}\right)\dfrac{\mathrm d\left(\sin\left(\frac yx\right)\right)}{\sin\left(\frac yx\right)}[/tex]
[tex]\mathrm df = \dfrac{\mathrm dx + \frac{\mathrm d(x^2+y^2)}{\sqrt{x^2+y^2}}}{x+\sqrt{x^2+y^2}}\ln\left(\sin\left(\dfrac yx\right)\right) + \ln\left(x+\sqrt{x^2+y^2}\right)\dfrac{\cos\left(\frac yx\right)\,\mathrm d\left(\frac yx\right)}{\sin\left(\frac yx\right)}[/tex]
[tex]\mathrm df = \dfrac{\mathrm dx + \frac{2x\,\mathrm dx+2y\,\mathrm dy}{\sqrt{x^2+y^2}}}{x+\sqrt{x^2+y^2}\right)\ln\left(\sin\left(\dfrac yx\right)\right) + \ln\left(x+\sqrt{x^2+y^2}}\right)\dfrac{\cos\left(\frac yx\right)\frac{x\,\mathrm dy-y\,\mathrm dx}{x^2}}{\sin\left(\frac yx\right)}[/tex]
[tex]\mathrm df = \dfrac{\left(2x+\sqrt{x^2+y^2}\right)\,\mathrm dx +2y\,\mathrm dy}{x\sqrt{x^2+y^2}+x^2+y^2\right)\ln\left(\sin\left(\dfrac yx\right)\right) \\\\ \indent + \dfrac1{x^2}\cot\left(\dfrac yx\right)\ln\left(x+\sqrt{x^2+y^2}}\right)(x\,\mathrm dy-y\,\mathrm dx)[/tex]
[tex]\mathrm df = \left(\left(\dfrac{2x+\sqrt{x^2+y^2}}{x\sqrt{x^2+y^2}+x^2+y^2}\right)\ln\left(\sin\left(\dfrac yx\right)\right) - \dfrac y{x^2}\cot\left(\dfrac yx\right)\ln\left(x+\sqrt{x^2+y^2}\right)\right)\,\mathrm dx \\\\ \indent + \left(\dfrac{2y}{x\sqrt{x^2+y^2}+x^2+y^2}\ln\left(\sin\left(\dfrac yx\right)\right)+\dfrac1x\cot\left(\dfrac yx\right)\ln\left(x+\sqrt{x^2+y^2}\right)\right)\,\mathrm dy[/tex]
